FIRST LAW OF THERMODYNAMICS 183

dharm

/M-therm/Th4-5.pm5

(ii)Turbine :

Power output of turbine :

Energy equation for turbine gives

mh 2 C^2 mh C

2

3 3

2

+ 22

F

HG

I

KJ

=+

F

HG

I

KJ

+ W2–3 [Q Q2–3 = 0, Z 1 = Z 2 ]

∴ Wmh

C mh C

23 2 2

2

3 3

2

− =+ 22

F

HG

I

KJ

−+

F

HG

I

KJ

=−+−

F

HG

I

KJ

L

N

M

M

O

Q

P

P

mh h() 23 CC^2

2

3

2

2

=−+−

L

N

M

M

O

Q

P

P

mc t tp() 23 CC^2

2

3

2

2

=−+−

×

L

N

M

M

O

Q

P

P

2 5 005 820 620^4055

2 1000

22

.( )1. () ()

= 2.5 [201 + 0.7125] = 504.3 kJ/s or 504.3 kW

Hence, power output of turbine = 504.3 kW. (Ans.)

(iii)Nozzle :

Velocity at exit from the nozzle :

Energy equation for nozzle gives,

h 3 + C^3

2

2 = h^4 +

C 42

2 [Q W3–4 = 0, Q3 – 4 = 0, Z^1 = Z^2 ]

C 42

2 = (h^3 – h^4 ) +

C 32

2 = cp(t^3 – t^4 ) +

C 32

2

= 1.005(620 – 510) + 2 1000^55

2

× = 112.062 × 10

(^3) J
∴ C 4 = 473.4 m/s.
Hence, velocity at exit from the nozzle = 473.4 m/s. (Ans.)

4.14. Heating-Cooling and Expansion of Vapours

The basic energy equations for non-flow and flow processes are also valid for vapours.

∴ When ∆KE = 0 and ∆PE = 0

dQ = du + p.dv ......for non-flow process.

dQ = dh – v.dp ......for flow process.

The various processes using vapour are discussed below :

1. Constant Volume Heating or Cooling. The constant volume heating process is repre-
   sented on p-v, T-s and h-s diagram as shown in Fig. 4.57 (a), (b), (c) respectively. It is assumed that
   the steam is in wet condition before heating at pressure p 1 , becomes superheated after heating and
   pressure increases from p 1 to p 2.
   Since the mass of steam, m, remains constant during the heating process,
   ∴ m = xvV vV

(^1) g 12

sup
, where V is the total constant volume of steam